Test Elements for Free Solvable Groups of Rank 2 | SpringerLink

Algebra and Logic, Vol. 40, No. 2, 2001

TEST ELEMENTS FOR FREE SOLVABLE GROUPS OF RANK 2 V. A. Roman’kov∗

UDC 512.54

It is proved that a free solvable group of rank 2 of degree 3 contains test elements. Thereby we solve the Fine–Shpilrain problem posed in [9, Question 14.88]. A family of elements g1 , . . . , gk of a group G is called a test collection if every endomorphism ϕ ∈ EndG for which ϕ(gi ) = gi (i = 1, . . . , k) is an automorphism. A test rank of G is the least cardinality of a test collection. A test element is a test collection consisting of one element. In [1] (see also [2]), it was proved that every endomorphism of a free group F2 of rank 2 with basis x, y which transforms the commutator (x, y) into its conjugate or inverse is an automorphism. Hence this commutator is a test element for F2 . A similar statement for a free metabelian group M2 of rank 2 was proven to hold in [3]. Every free group Fn of finite rank n possesses test elements and, consequently, has test rank 1. This has been made obvious by the results of Zieschang, Rosenberger, and Rips (see also Shpilrain [4, 5] and Turner [6]; the latter gives a description of test elements for free groups of finite ranks). In [7], it was proved that all non-identity elements of a commutant of the free metabelian group M2 of rank 2 are test elements for M2 . Moreover, there a free metabelian group Mr of finite rank r was shown to have its test rank equal to r − 1 and a description was furnished for all minimal test collections of Mr . For the case of a free solvable group Srl of rank r of solvability index l > 3, the situation remained unclear. In [8], it was shown that the commutator of basis elements for r = 2 is no longer a test element. In [9, Question 14.88], Fine and Shpilrain articulated the problem which inquires whether there exist test elements for free solvable groups S2l , l > 3 (see also the Magnus program web site http://www.grouptheory.org, “Open problems,” Question S9). The goal of the present article is seeking a test element for S = S23 . We prove the following: THEOREM. A free solvable group S of rank 2 of solvability index 3 contains test elements each of which belongs to a last non-identity commutant of the group. Thus the answer to the above-cited problem is positive for at least solvability index 3. For commutation, commutators, and simple commutators of elements of an arbitrary group G, we adopt the following notation: g f = f gf −1 , (f, g) = f gf −1 g −1 , (f1 , . . . , fn+1 ) = ((f1 , . . . , fn ), fn+1 ).

(1)

As usual, by G0 we denote a commutant, and by G(k) the kth commutant of G. Elements of the lower central series are denoted by γi G, i = 1, 2, . . . . ∗ Supported

by RFFR grant No. 98-01-00932 and by the RF Ministry of Education.

Translated from Algebra i Logika, Vol. 40, No. 2, pp. 192-201, March-April, 2001. Original article submitted March 21, 2000.

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c 2001 Plenum Publishing Corporation 0002-5232/01/4002-0106 $25.00

We need some facts concerning metabelian groups (see, e.g., [10]). Every metabelian group satisfies all identities of the form (f, g, h1 , . . . , hk ) = (f, g, hσ(1) , . . . , hσ(k) ),

(2)

where σ is an arbitrary permutation on {1, . . . , k}. Let M = M2 be a free metabelian group of rank 2 with basis x, y. Basis commutators which allow the elements of M to be uniquely represented modulo γi M are elements such as x, y, z = z(1, 1) = (y, x), z(k, l) = (y, x, x, . . . , x, y, y, . . . , y), k + l 6 i − 1, where k and l stand for, respectively, the number of occurrences of x and the number of occurrences of y. Basis commutators are naturally ordered primarily in increasing weight, and secondarily, in increasing the number of occurrences of y. Every element g of M is uniquely represented thus: Y g ≡ xα y β z(k, l)λ(k,l) (mod γi M ) , α, β, λ(k, l) ∈ Z, (3) k+l6i−1

where basis commutators in the product are arranged in ascending order. In particular, the image of every element z(k0 , l0 ) (k0 + l0 = i − 1) w.r.t. an endomorphism ϕ ∈ EndM is uniquely represented modulo γi M as a product of degrees of basis commutators like z(k, l) (k + l = i − 1). We may assume that they are arranged in ascending order. Now consider a free group T of rank 2 in the variety of all groups with nilpotent commutant of class 2. Assume that M = T /T (2) and that its basis x, y agrees with a basis x1 , y1 for T . Denote by z1 (k, l) a commutator obtained from z(k, l) by replacing x and y by x1 and y1 , respectively. For the resulting commutators, we keep the former order and conceive of them as simple basis commutators for T . In addition, for T we define commutators w1 (k, l, m, n) = (z1 (k, l), z1 (m, n)). Among them are basis commutators distinguished by the extra condition that z1 (k, l) > z1 (m, n). The basis commutators of T are arranged in order of increasing weight, and commutators of a same weight — in order of increasing the number of occurrences of y1 ; in other cases, the order is given arbitrarily. Note that T (2) is a module over a group ring ZA of a free Abelian group A = M/M 0 = T /T 0 with respective basis x0 , y0 . We know from [11] that every element of T is uniquely representable modulo γi T as a product of degrees of basis commutators of weight at most i − 1, arranged in ascending order. Define the element u1 ∈ T by setting u1 = w1 (3, 2, 1, 1)w1 (2, 2, 1, 2).

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In [12], it was stated that if an image of the element z(2, 2) w.r.t. ϕ ∈ EndM is comparable with z(2, 2)±1 modulo γ5 M then ϕ(x) ≡ xε and ϕ(y) ≡ y η (mod γ2 M ), ε, η ∈ {1, −1}. This entails the following: LEMMA. Let an image of u1 w.r.t. ϕ ∈ EndT be comparable with u1 modulo γ8 T . Then ϕ(x1 ) ≡ x1 and ϕ(y1 ) ≡ y1 (mod γ2 T ). Proof. Assume ϕ1 (u1 ) = u1 for ϕ1 ∈ EndT . Let ϕ1 induce ϕ from EndM . It is then easy to see that the image of u1 w.r.t. ϕ1 is uniquely determined from ϕ. Let ϕ be given by the map x 7→ (y, x)α(1−x0 )+β(y0 ) x, y 7→ (y, x)γ y,

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where α, β(y0 ), γ ∈ ZA and β(y0 ) depends on y0 only.

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An element (y, x)α(1−x0 ) in T has a preimage representable as v1 = (w1 , x1 ), w1 ∈ T 0 . We modify ϕ1 by multiplying it from the right by an inner automorphism of T corresponding to the element w1−1 . This yields an endomorphism for which u1 still is a fixed element, and the induced endomorphism of M is defined by the map x 7→ (y, x)β(y0 ) x, y 7→ (y, x)δ y,

(6)

where δ ∈ ZA. For simplicity, we preserve the notation for ϕ1 and for its induced endomorphism ϕ. The image of u1 still depends on just ϕ. Let (y, x)β(y0 ) ∈ γk M \ γk+1 M , (y, x)δ ∈ γl M \ γl+1 M . First assume that k 6= l. Let k < l [for l < k (or β(y0 ) = 0), computations are similar]. Then (y, x)β(y0 ) ≡ z(1, k − 1)t (mod γk+1 M ) , t 6= 0.

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Compute ϕ1 (u1 ) (mod γ7+k T ) as follows: ϕ1 (u1 ) ≡ ((y1 , z1 (1, k − 1)t x1 , x1 , x1 , y1 ), (y1 , z1 (1, k − 1)t x1 )) ((y1 , z1 (1, k − 1)t x1 , x1 , y1 ), (y1 , z1 (1, k − 1)t x1 , y1 ) ≡ w1 (3, k + 1, 1, 1)−t w1 (3, 2, 1, k)−t w1 (3, 2, 1, 1)w1 (2, k + 1, 1, 2)−t w1 (2, 2, 1, k + 1)−t w1 (2, 2, 1, 2) (mod γk+7 T ).

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The same argument applies with δ = 0. We arrive at a contradiction, for the right part in (8) is a product of non-zero degrees of the basis commutators incomparable with u1 . If, however, k = l, and if we put (y, x)β(y0 ) ≡ z(1, k − 1)t (mod γk+1 M ), t 6= 0, as above in doing computations, then we see that the degree of a basis commutator w1 (3, k + 1, 1, 1)−t fails to cancel, since the basis commutators “coming” from y1 and including z1 (1, 1) have strictly more than three occurrences of x1 in the other part. This is a contradiction again. The only possibility we are left with is an endomorphism ϕ1 ∈ EndT acting identically modulo T (2) . Let it be defined by the map x1 7→ c1 x1 , y1 7→ d1 y1 , c1 , d1 ∈ T (2) .

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Then ϕ1 acts identically on T (2) . It is easy to verify that a similar map, used with c−1 and d−1 1 1 , defines −1 ϕ1 , that is, ϕ1 ∈ Aut T . Hence u1 is a test element for T . We pass directly to: Proof of the theorem. Let T = S/γ2 S 0 . Suppose that x2 , y2 is a basis for S corresponding to the basis x1 , y1 . We claim that an element u2 obtained from u1 by replacing the basis elements x1 and y1 by x2 and y2 , respectively, is a test element for S. Let ϕ2 ∈ EndS be an endomorphism such that ϕ2 (u2 ) = u2 . It induces an endomorphism ϕ1 ∈ EndT for which u1 is a fixed element. Multiplying ϕ2 by a suitable inner automorphism of S, in view of the above, we arrive in the following situation (preserving the notation for simplicity): the endomorphism ϕ2 0 induces an identity map on M = S/S (2) and transforms u2 into its conjugate uw 2 , where w ∈ S . To make our discussion self-contained, we give an argument which basically is a replica of the argument in [13]. Let ϕ2 be defined by the map x2 7→ cx2 , y2 7→ dy2 , c, d ∈ S (2) .

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We know from [14, 15] that the group ring ZM is embeddable in a division ring P of formal series over Q. At the moment we look into some concepts which the proof of the theorem requires to be brought in. Every element of M is uniquely represented as xi g, where g ∈ gr(y, M 0 ), i ∈ Z. If the element γ ∈ P has the form γ = xi0 a0 + . . . + xi0 +k ak , a0 , ak 6= 0,

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where elements ai all belong to a subdivision ring of P generated by gr(y, M 0 ), then we say that γ has a finite decomposition in terms of x. For such γ’s, define an integer-valued additive function ρx (γ) = k. If ρx (γ) = 0 then γ is said to be homogeneous in x. The function ρy can be defined similarly. In [13], note, these concepts were introduced in a much more general setting — in particular, for free solvable groups of any rank and arbitrary degree. In [13, Lemma 3], it was proved that if a is a non-zero element of P , γ has a finite decomposition in terms of x, and for all k ∈ N, aγ −k ∈ ZM , then ρx (γ) = 0 and γ is homogeneous in x. Again, in [13, Lemma 4] is stated that if ab−k ∈ ZM for any k ∈ N and for non-zero elements a, b ∈ ZM then either b ∈ M or −b ∈ M . A module S (2) over the group ring ZM is naturally embeddable in the vector space V over a division ring P of ZM of dimension 1 (cf. [13]). We use partial Fox derivatives D1 = ∂/∂x2 and D2 = ∂/∂y2 which are defined on S with values in ZM (for their requisite properties, see, e.g., [13]). Since ϕ2 is identical modulo S (2) , its action extends to the whole V . Thus if we replace w ∈ S 0 by its image in M (preserving the notation for simplicity), we arrive at ϕ2 (u2 ) = uw 2 , w ∈ M.

(12)

We proceed to prove that w = 1. Differentiating (12) yields wD1 u2 = D1 ϕ2 (u2 ) = D1 u2 D1 c + D1 u2 + D2 u2 D1 d.

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Let wD1 u2 = D1 u2 α, α ∈ P ; then wl D1 u2 = D1 u2 αl ∈ ZM

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for all l ∈ N. Moreover, this is true for all l ∈ Z since w is an element of M . The basic relation for Fox derivatives in [13] delivers D2 u2 = D1 u2 (x − 1)(1 − y)−1 .

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α = D1 c + 1 + (x − 1)(1 − y)−1 D1 d.

(16)

Hence

Clearly, α has a finite decomposition in terms of x. In view of the remarks above, α is homogeneous in x. We claim that this is possible only if α ∈ M or −α ∈ M . Let D1 d = y i0 a0 + . . . + y i0 +k ak , a0 , ak 6= 0, be a decomposition of the element D1 d in terms of y. And write it in the form D1 d = (y i0 − 1)a0 + . . . + (y i0 +k − 1)ak + a,

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where a = a0 + . . . + ak . Keeping in mind that y i − 1 is divisible by y − 1 in the ring Z(gr(y)), we obtain α = r + (x − 1)(1 − y)−1 a,

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where r, a ∈ ZM and a is homogeneous in y. We claim that a = 0, and so α ∈ ZM . To prove this, let a be decomposed in terms of x as follows: a = xj0 b0 + . . . + xj0 +t bt , b0 , bt 6= 0.

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We compute coefficients at xj0 and xj0 +t+1 in the decomposition of (x − 1)(1 − y)−1 a in terms of x. Since (1 − y)−1 = 1 + y + y 2 + . . . + y k + . . . ,

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−(1 + y z + y 2z + . . . + y kz + . . . )b0 , z = x−j0 .

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the coefficient at xj0 equals

The coefficient at xj0 +t+1 is computed via the same formula, but z = x−j0 +t and the parenthesized expression is multiplied by bt instead of b0 . If a 6= 0 then the coefficients above do not belong to ZM since a is homogeneous in y. Therefore α is not homogeneous in x, a contradiction. Hence a = 0, so α ∈ ZM , and by the remark made above, α = εg1 , g1 ∈ M, ε ∈ {1, −1}.

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Consequently, (1 − y)−1 D1 d ∈ ZM . Applying a trivialization to (16) yields ε = 1. Let wD2 u2 = D2 u2 β for some β ∈ P . Similarly to the above we prove that β = g2 ∈ M . Then D1 u2 g1 (x − 1) + D2 u2 g2 (y − 1) = 0.

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g1 (x − 1) + (x − 1)(1 − y)−1 g2 (y − 1) = 0.

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Hence

This equality holds only if yg2 = g2 y. Since M is not Abelian, by the Mal’tsev theorem in [16] it follows that g2 = y i , i ∈ Z. That g1 = xj , j ∈ Z, can be proved similarly. Substituting these values in the previous equality delivers i = j = 0; so, ϕ2 (u2 ) = u2 . Now, since S (2) is one-dimensional over P and ϕ2 is trivial modulo S (2) , ϕ acts identically on S (2) . In this case ϕ2 (S) = S, and since S is Hopf, ϕ2 ∈ AutS. An arbitrary test element for S will belong to a subgroup S (2) in virtue of the following: (l−1)

Proposition. If g is a test element for S2l , l > 2, then g ∈ S2l . Proof. For l = 2, the statement is proved in [7]. A proof for the general case is by and large its replica. (l−1) Let g 6∈ S2l be a test element and D1 and D2 be partial Fox derivatives w.r.t. basis elements x and y of S2l with values in ZS2l−1 . An endomorphism ϕ ∈ EndS2l given by the map x 7→ uλ x, y 7→ uµ y, (l−1)

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where 1 6= u ∈ S2l and D1 gλ + D2 gµ = 0, keeps an element g fixed. Since the ring ZS2l−1 satisfies the Ore condition, there exist λ and µ which satisfy the above equality and are not both equal to zero together.

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Note that ϕ(u) 6= u. In fact, ϕ(u) = uD1 uλ+D2 uµ+1 .

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For simplicity, let images of elements of S2l in S2l−1 preserve the notation used for the elements proper. Using the basic identity for derivatives yields D1 u(x−1) + D2 u(y − 1) = 0, and so λ = (x− 1)ν, µ = (y − 1)ν for some 0 6= ν ∈ P . It follows that D1 gλ + D2 gµ = (g − 1)ν 6= 0, which contradicts the assumption. (l−1) The automorphism ϕ acts identically modulo S2l and acts non-identically on u, which is impossible by [13]. We have thus proved the proposition, completing the theorem. Remark. We are unaware whether every non-identity element of S (2) is a test element for S (2) as treated by analogy with M2 . The question as to whether there exist test elements for S2l , l > 4, also remains open. REFERENCES 1. J. Nielsen, “Die Isomorphismen der allgemeinen unendlichen Gruppe mit zwei Erzeugenden,” Math. Ann., No. 78, 269-272 (1918). 2. R. C. Lyndon and P. E. Schupp, Combinatorial Group Theory, Springer, Berlin (1977). 3. V. G. Durnev, “Free generators for a free metabelian group of rank 2,” Dep. VINITI, No. 4036-B87 (1987). 4. V. Shpilrain, “Recognizing automorphisms of the free groups,” Arch. Math., 62, No. 5, 385-392 (1994). 5. V. Shpilrain, “Test elements for endomorphisms of free groups and algebras,” Isr. J. Math., 92, Nos. 1-3, 307-316 (1995). 6. E. Turner, “Test words for automorphisms of free groups,” Bull. Lond. Math. Soc., 28, No. 3, 255-263 (1996). 7. E. I. Timoshenko, “Test elements and test rank for a free metabelian group,” Sib. Mat. Zh., 41, No. 6, 1451-1456 (2000). 8. N. Gupta and V. Shpilrain, “Nielsen’s commutator test for two-generator groups,” Math. Proc. Camb. Phil. Soc., 114, No. 2, 295-301 (1993). 9. The Kourovka Notebook, 14th edn., Institute of Mathematics SO RAN, Novosibirsk (1999). 10. H. Neumann, Varieties of Groups, Springer, Berlin (1967). 11. A. L. Shmel’kin, “Free polynilpotent groups,” Dokl. Akad. Nauk SSSR, 169, No. 5, 1024-1025 (1967). 12. V. A. Roman’kov, “Equations in free metabelian groups,” Sib. Mat. Zh., 20, No. 3, 671-673 (1979). 13. V. A. Roman’kov, “Normal automorphisms of discrete groups,” Sib. Mat. Zh., 24, No. 4, 138-149 (1983). 14. A. I. Mal’tsev, “Embeddings of group algebras in division algebras,” Dokl. Akad. Nauk SSSR, 60, No. 9, 1499-1501 (1948). 15. B. H. Neumann, “On ordered division rings,” Trans. Am. Math. Soc., 66, No. 1, 202-252 (1949). 16. A. I. Mal’tsev, “Free solvable groups,” Dokl. Akad. Nauk SSSR, 130, No. 3, 495-498 (1960).

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